Question

在我的表实习中，我有两个外键 id_promoter_internship 和 id_supervisor_internship 。在表business_contacts中，我有1个主键 id_business 。我正在尝试从business_contacts表中获取数据，该表链接到实习表。以下查询是否正确？

public function update_form_business_contact($name_enterprise){
    $query = "
        SELECT 
        * 
        FROM business_contacts
            ,internships 
        WHERE 
            business_contacts.id_business = internships.id_supervisor_internship 
            AND internships.name_enterprise_internship = '$name_enterprise'";       
    $result = $this->_db->query($query);

    # Go through results of teachers
    if($result->rowCount()!=0){
        while($row=$result->fetch()){
            $contact= new businesscontact ( $row->id_business,$row->firstname_business,$row->lastname_business,$row->service_business,$row->function_business,$row->phone_business,$row->phone_secretary_business,$row->mobile_business);
        }
    }
    return $contact;
}

我的问题是：主键是否必须引用两个外键？如果是这样，我该怎么做呢。谢谢你的帮助。

Answer 1

你应该使用JOIN（因为那是FK的用途）。 JOIN语法告诉DB如何关联表（并使其他任何处理代码的人都清楚）。顺便提一句，您应该避免SELECT *并列出您想要的字段。

SELECT bc.*, i.* 
FROM 
    business_contacts bc
        INNER JOIN internships i ON bc.id_business = i.id_supervisor_internship 
WHERE 
    i.name_enterprise_internship = '$name_enterprise'

您只需要引用相关的字段。在您写入数据时强制执行FK。

Answer 2

您正试图从internships获取数据，因此启动：

SELECT *
FROM internships

这为您提供实习机会，其中包括联系人的标识符。要为这些联系人添加信息，请加入包含该信息的表。在这种情况下，有两个连接，每个联系一个：

SELECT *
FROM internships
INNER JOIN business_contacts promoter
  ON internships.id_promoter_internship = promoter.id
INNER JOIN business_contacts supervisor
  ON internships.id_supervisor_internship = supervisor.id

通过提供表别名（在这种情况下为promoter和supervisor），您可以轻松地连接同一个表两次（或者您想要的次数，真的）并将其视为单独的表

结果是internships中的所有记录以及business_contacts中与internships记录中每个键相关的其他数据。

您还可以对列进行别名，以使结果更加明确。类似的东西：

SELECT
    internships.some_field AS some_field,
    promoter.some_field AS promoter_some_field,
    supervisor.some_field AS supervisor_some_field
FROM
...

Answer 3

不是很糟糕，但你可能想试试这个：

    <?php

    public function update_form_business_contact($name_enterprise){
        $query = "
                 SELECT
                      *
                 FROM business_contacts bCon
                 INNER JOIN internships  iShip
                 ON bCon.id=iShip.id_supervisor_internship
                 WHERE
                     iShip.name_enterprise_internship = '$name_enterprise'";
        $result = $this->_db->query($query);

        # Go through results of teachers
        if($result->rowCount()!=0){
            while($row=$result->fetch()){
                $contact= new businesscontact ( $row->id_business,$row->firstname_business,$row->lastname_business,$row->service_business,$row->function_business,$row->phone_business,$row->phone_secretary_business,$row->mobile_business);
            }
        }
        return $contact;
    }

关于2个FK和1个主键

3 个答案: